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Mean Value Theorem - AP Calculus AB

  1. A function gg is continuous on [2,10][2,10] with g(2)=3g(2)=3 and g(10)=−5g(10)=-5. Does the Mean Value Theorem guarantee a cc with g′(c)=−1g'(c)=-1?
    Answer: Yes, because gg is continuous on [2,10][2,10] and differentiable on (2,10)(2,10), so some cc has g′(c)=−5−310−2=−1g'(c)=\dfrac{-5-3}{10-2}=-1. The Mean Value Theorem requires g to be continuous on [2,10] and differentiable on (2,10). Since these conditions are satisfied, MVT guarantees there exists c in (2,10) where g'(c) equals the average rate of change. Computing: (g(10)-g(2))/(10-2) = (-5-3)/8 = -8/8 = -1. Therefore, MVT does guarantee a point where g'(c) = -1. A common error is thinking MVT cannot apply when functions become negative, but the theorem works regardless of the signs of function values. Only continuity and differentiability conditions restrict MVT's application.
  2. A function FF is continuous on [3,7][3,7] with F(3)=9F(3)=9 and F(7)=1F(7)=1. Does the Mean Value Theorem guarantee a cc with F′(c)=−2F'(c)=-2?
    Answer: Yes, because FF is continuous on [3,7][3,7] and differentiable on (3,7)(3,7), so some cc has F′(c)=1−97−3=−2F'(c)=\dfrac{1-9}{7-3}=-2. The Mean Value Theorem applies when F is continuous on [3,7] and differentiable on (3,7). Since both conditions are satisfied, MVT guarantees there exists c in (3,7) where F'(c) equals the average rate of change. The average rate of change is (F(7)-F(3))/(7-3) = (1-9)/4 = -8/4 = -2. Therefore, MVT does guarantee a point where F'(c) = -2. Students sometimes incorrectly think derivatives cannot be negative, but derivatives can be any real number. A negative derivative simply indicates the function is decreasing at that point. MVT makes no restrictions on the sign of the guaranteed derivative value.
  3. A function mm is continuous on [0,12][0,12] with m(0)=6m(0)=6 and m(12)=0m(12)=0. Does the Mean Value Theorem guarantee a cc with m′(c)=−12m'(c)=-\tfrac{1}{2}?
    Answer: Yes, because mm is continuous on [0,12][0,12] and differentiable on (0,12)(0,12), so some cc has m′(c)=0−612−0=−12m'(c)=\dfrac{0-6}{12-0}=-\tfrac{1}{2}. The Mean Value Theorem applies when m is continuous on [0,12] and differentiable on (0,12). Since both conditions are satisfied, MVT guarantees there exists c in (0,12) where m'(c) equals the average rate of change. The average rate of change is (m(12)-m(0))/(12-0) = (0-6)/12 = -6/12 = -1/2. Therefore, MVT does guarantee a point where m'(c) = -1/2. Students might incorrectly think MVT requires integer slopes, but the theorem works for any real-valued average rate of change. Fractional derivatives are perfectly valid and commonly occur in MVT applications.
  4. A function XX is continuous on [2,3][2,3] with X(2)=4X(2)=4 and X(3)=1X(3)=1. Does the Mean Value Theorem guarantee a cc with X′(c)=−3X'(c)=-3?
    Answer: Yes, because XX is continuous on [2,3][2,3] and differentiable on (2,3)(2,3), so some cc has X′(c)=1−43−2=−3X'(c)=\dfrac{1-4}{3-2}=-3. The Mean Value Theorem applies when X is continuous on [2,3] and differentiable on (2,3). Since both conditions are satisfied, MVT guarantees there exists c in (2,3) where X'(c) equals the average rate of change. The average rate of change is (X(3)-X(2))/(3-2) = (1-4)/1 = -3/1 = -3. Therefore, MVT does guarantee a point where X'(c) = -3. Students might think MVT cannot apply on intervals of length 1, but interval length doesn't affect the theorem's validity. As long as the function satisfies continuity and differentiability conditions, MVT applies regardless of how short or long the interval is.
  5. A function hh is continuous on [2,6][2,6] and differentiable on (2,6)(2,6) with h(2)=−3h(2)=-3 and h(6)=5h(6)=5. Does MVT guarantee cc with h′(c)=2h'(c)=2?
    Answer: Yes, because hh is continuous on [2,6][2,6], differentiable on (2,6)(2,6), and h(6)−h(2)6−2=5−(−3)4=2\frac{h(6)-h(2)}{6-2}=\frac{5-(-3)}{4}=2. To apply the Mean Value Theorem, we need continuity on [2,6] and differentiability on (2,6), both of which are given. The average rate of change over this interval is (h(6)-h(2))/(6-2) = (5-(-3))/4 = 8/4 = 2. Since this average rate equals the desired derivative value of 2, MVT guarantees there exists at least one c in (2,6) where h'(c) = 2. The fact that h changes from negative to positive values is irrelevant to MVT's application. A common error is confusing MVT with the Intermediate Value Theorem, which deals with function values rather than derivatives. Remember: MVT connects the average rate of change to an instantaneous rate of change somewhere in the interval.
  6. A function aa is continuous on [0,7][0,7] with a(0)=1a(0)=1 and a(7)=15a(7)=15. Does the Mean Value Theorem guarantee a cc with a′(c)=2a'(c)=2?
    Answer: Yes, because aa is continuous on [0,7][0,7] and differentiable on (0,7)(0,7), so some cc has a′(c)=15−17−0=2a'(c)=\dfrac{15-1}{7-0}=2. The Mean Value Theorem requires a to be continuous on [0,7] and differentiable on (0,7). Since these conditions are satisfied, MVT guarantees there exists c in (0,7) where a'(c) equals the average rate of change. Computing: (a(7)-a(0))/(7-0) = (15-1)/7 = 14/7 = 2. Therefore, MVT does guarantee a point where a'(c) = 2. A common misconception is that the derivative value being "too small" prevents MVT application, but MVT makes no restrictions on the magnitude of the guaranteed derivative. The theorem works for any derivative value that equals the computed average rate of change.
  7. A function tt is continuous on [−4,2][-4,2] with t(−4)=9t(-4)=9 and t(2)=3t(2)=3. Does the Mean Value Theorem guarantee a cc with t′(c)=−1t'(c)=-1?
    Answer: Yes, because tt is continuous on [−4,2][-4,2] and differentiable on (−4,2)(-4,2), so some cc has t′(c)=3−92−(−4)=−1t'(c)=\dfrac{3-9}{2-(-4)}=-1. The Mean Value Theorem applies when t is continuous on [-4,2] and differentiable on (-4,2). Since both conditions are met, MVT guarantees there exists c in (-4,2) where t'(c) equals the average rate of change. The average rate of change is t(2)−t(−4)2−(−4)=3−96=−66=−1\frac{t(2)-t(-4)}{2-(-4)} = \frac{3-9}{6} = \frac{-6}{6} = -1. Therefore, MVT does guarantee a point where t'(c) = -1. A common error is thinking MVT requires the function to be increasing, but the theorem works for both increasing and decreasing functions. The negative average rate of change simply indicates the function decreases overall on the interval.
  8. A function ff is continuous on [2,8][2,8] with f(2)=5f(2)=5 and f(8)=5f(8)=5. Does the Mean Value Theorem guarantee a cc with f′(c)=0f'(c)=0?
    Answer: Yes, because f(2)=f(8)f(2)=f(8), so the average rate of change is 0 and MVT gives f′(c)=0f'(c)=0 for some cc. The Mean Value Theorem applies when f is continuous on [2,8] and differentiable on (2,8). Since both conditions are met, MVT guarantees there exists c in (2,8) where f'(c) equals the average rate of change. The average rate of change is (f(8)-f(2))/(8-2) = (5-5)/6 = 0. Therefore, MVT does guarantee a point where f'(c) = 0. This is actually a special case leading toward Rolle's Theorem - when endpoints are equal, the average rate of change is zero, so MVT guarantees at least one point with zero derivative. Students should recognize that equal endpoints make MVT application straightforward and powerful.
  9. A car's position s(t)s(t) is continuous on [0,8][0,8] and differentiable on (0,8)(0,8) with s(0)=5s(0)=5 and s(8)=21s(8)=21. Does MVT guarantee a time cc with s′(c)=2s'(c)=2?​
    Answer: Yes, because ss is continuous on [0,8][0,8] and differentiable on (0,8)(0,8), and 21−58−0=2\frac{21-5}{8-0}=2. In this physics context, s(t) represents position at time t, and s'(t) represents velocity. Since s is continuous on [0,8] and differentiable on (0,8), MVT applies. The average velocity over the interval is (s(8)-s(0))/(8-0) = (21-5)/8 = 16/8 = 2. MVT guarantees there's at least one moment c in (0,8) where the instantaneous velocity s'(c) equals this average velocity of 2. This makes physical sense: if you travel 16 units in 8 hours, at some point your speedometer must read exactly 2 units per hour. The key insight: MVT connects average rates (like average velocity) to instantaneous rates (like instantaneous velocity).
  10. A function ww is continuous on [−1,1][-1,1] with w(−1)=2w(-1)=2 and w(1)=6w(1)=6. Does the Mean Value Theorem guarantee a cc with w′(c)=2w'(c)=2?
    Answer: Yes, because ww is continuous on [−1,1][-1,1] and differentiable on (−1,1)(-1,1), so some cc has w′(c)=6−21−(−1)=2w'(c)=\dfrac{6-2}{1-(-1)}=2. The Mean Value Theorem requires w to be continuous on [-1,1] and differentiable on (-1,1). Since these conditions are met, MVT guarantees there exists c in (-1,1) where w'(c) equals the average rate of change. The average rate of change is (w(1)-w(-1))/(1-(-1)) = (6-2)/2 = 4/2 = 2. Therefore, MVT does guarantee a point where w'(c) = 2. Students might think interval length matters for MVT application, but the theorem works on any interval where the function satisfies the required conditions. Whether the interval is long or short doesn't affect MVT's validity - only continuity and differentiability matter.
  11. A function JJ is continuous on [−5,−1][-5,-1] with J(−5)=6J(-5)=6 and J(−1)=2J(-1)=2. Does the Mean Value Theorem guarantee a cc with J′(c)=−1J'(c)=-1?
    Answer: Yes, because JJ is continuous on [−5,−1][-5,-1] and differentiable on (−5,−1)(-5,-1), so some cc has J′(c)=2−6−1−(−5)=−1J'(c)=\dfrac{2-6}{-1-(-5)}=-1. The Mean Value Theorem requires J to be continuous on [-5,-1] and differentiable on (-5,-1). Since these conditions are satisfied, MVT guarantees there exists c in (-5,-1) where J'(c) equals the average rate of change. Computing: (J(-1)-J(-5))/(-1-(-5)) = (2-6)/4 = -4/4 = -1. Therefore, MVT does guarantee a point where J'(c) = -1. A common misconception is that negative x-values prevent MVT application, but the theorem works on any interval where continuity and differentiability conditions are met. The location of the interval on the x-axis is irrelevant to MVT's validity.
  12. A function GG is continuous on [2,6][2,6] with G(2)=−1G(2)=-1 and G(6)=7G(6)=7. Does the Mean Value Theorem guarantee a cc with G′(c)=2G'(c)=2?
    Answer: Yes, because GG is continuous on [2,6][2,6] and differentiable on (2,6)(2,6), so some cc has G′(c)=7−(−1)6−2=2G'(c)=\dfrac{7-(-1)}{6-2}=2. The Mean Value Theorem requires G to be continuous on [2,6] and differentiable on (2,6). Since these conditions are satisfied, MVT guarantees there exists c in (2,6) where G'(c) equals the average rate of change. Computing: (G(6)-G(2))/(6-2) = (7-(-1))/4 = 8/4 = 2. Therefore, MVT does guarantee a point where G'(c) = 2. A common error is thinking negative function values prevent MVT application, but the theorem works regardless of whether function values are positive, negative, or mixed. The key is meeting the continuity and differentiability requirements, not the signs of the function values.
  13. A function ff is continuous on [1,5][1,5] with f(1)=2f(1)=2 and f(5)=10f(5)=10. Does the Mean Value Theorem guarantee a cc with f′(c)=2f'(c)=2?
    Answer: Yes, because ff is continuous on [1,5][1,5] and differentiable on (1,5)(1,5), so some cc has f′(c)=10−25−1=2f'(c)=\dfrac{10-2}{5-1}=2. The Mean Value Theorem applies when a function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b). Since f satisfies both conditions on [1,5], MVT guarantees there exists some c in (1,5) where f'(c) equals the average rate of change. The average rate of change is (f(5)-f(1))/(5-1) = (10-2)/(4) = 2. Therefore, MVT does guarantee a point c where f'(c) = 2. A common error is thinking that knowing only endpoint values is insufficient - but when continuity and differentiability are given, MVT provides this powerful conclusion. Always check both MVT conditions before applying the theorem.
  14. A function ee is continuous on [2,7][2,7] with e(2)=9e(2)=9 and e(7)=4e(7)=4. Does the Mean Value Theorem guarantee a cc with e′(c)=−1e'(c)=-1?
    Answer: Yes, because ee is continuous on [2,7][2,7] and differentiable on (2,7)(2,7), so some cc has e′(c)=4−97−2=−1e'(c)=\dfrac{4-9}{7-2}=-1. The Mean Value Theorem requires e to be continuous on [2,7] and differentiable on (2,7). Since these conditions are met, MVT guarantees there exists c in (2,7) where e'(c) equals the average rate of change. The average rate of change is (e(7)-e(2))/(7-2) = (4-9)/5 = -5/5 = -1. Therefore, MVT does guarantee a point where e'(c) = -1. Students sometimes think the derivative value must be "between" the endpoint function values, but MVT guarantees the derivative equals the average rate of change, which can be any real number depending on the endpoint values and interval length.
  15. A function ff is continuous on [0,2][0,2] and differentiable on (0,2)(0,2) with f(0)=1f(0)=1 and f(2)=1f(2)=1. Does MVT guarantee some cc with f′(c)=0f'(c)=0?
    Answer: Yes, because ff is continuous on [0,2][0,2], differentiable on (0,2)(0,2), and f(2)−f(0)2−0=0\frac{f(2)-f(0)}{2-0}=0. This is a special case of MVT known as Rolle's Theorem, which applies when f(a) = f(b). Here, f is continuous on [0,2] and differentiable on (0,2), with f(0) = f(2) = 1. The average rate of change is (f(2)-f(0))/(2-0) = (1-1)/2 = 0/2 = 0. Since MVT guarantees a point where the derivative equals this average rate, there must exist at least one c in (0,2) where f'(c) = 0. This doesn't mean the derivative is zero everywhere (choice A's error), just at least at one point. A common misconception is that f'(c) = 0 requires a maximum or minimum, but it could also occur at an inflection point. When endpoints have equal function values, MVT simplifies to guaranteeing a horizontal tangent somewhere.
  16. A function uu is continuous on [1,9][1,9] with u(1)=0u(1)=0 and u(9)=16u(9)=16. Does the Mean Value Theorem guarantee a cc with u′(c)=2u'(c)=2?
    Answer: Yes, because uu is continuous on [1,9][1,9] and differentiable on (1,9)(1,9), so some cc has u′(c)=16−09−1=2u'(c)=\dfrac{16-0}{9-1}=2. The Mean Value Theorem requires u to be continuous on [1,9] and differentiable on (1,9). Since these conditions are satisfied, MVT guarantees there exists c in (1,9) where u'(c) equals the average rate of change. Computing: (u(9)-u(1))/(9-1) = (16-0)/8 = 16/8 = 2. Therefore, MVT does guarantee a point where u'(c) = 2. Students sometimes confuse the change in function values u(9)-u(1) = 16 with the derivative value, but MVT specifically guarantees the derivative equals the average rate of change, which requires dividing by the interval length. Always compute (f(b)-f(a))/(b-a) for MVT applications.
  17. A function rr is continuous on [1,5][1,5] with r(1)=12r(1)=12 and r(5)=4r(5)=4. Does the Mean Value Theorem guarantee a cc with r′(c)=−2r'(c)=-2?
    Answer: Yes, because rr is continuous on [1,5][1,5] and differentiable on (1,5)(1,5), so some cc has r′(c)=4−125−1=−2r'(c)=\dfrac{4-12}{5-1}=-2. The Mean Value Theorem requires r to be continuous on [1,5] and differentiable on (1,5). Since these conditions are satisfied, MVT guarantees there exists c in (1,5) where r'(c) equals the average rate of change. Computing: (r(5)-r(1))/(5-1) = (4-12)/4 = -8/4 = -2. Therefore, MVT does guarantee a point where r'(c) = -2. A common misconception is that negative slopes are not guaranteed by MVT, but the theorem makes no restrictions on the sign of the derivative. Whether positive or negative, the derivative equals the average rate of change.
  18. Given ff is continuous on [−2,4][-2,4] and differentiable on (−2,4)(-2,4) with f(−2)=3f(-2)=3 and f(4)=−9f(4)=-9, does MVT guarantee cc where f′(c)=−2f'(c)=-2?​
    Answer: Yes, because ff is continuous on [−2,4][-2,4] and differentiable on (−2,4)(-2,4), so some cc has f′(c)=−9−34−(−2)=−2f'(c)=\frac{-9-3}{4-(-2)}=-2. This problem provides both conditions needed for the Mean Value Theorem: f is continuous on [-2,4] and differentiable on (-2,4). The MVT states that there exists at least one c in (-2,4) where f'(c) equals the average rate of change: (f(4)-f(-2))/(4-(-2)) = (-9-3)/6 = -12/6 = -2. The fact that f(-2) ≠ f(4) is irrelevant; that condition is only needed for Rolle's Theorem (a special case of MVT). A common misconception is thinking MVT requires equal endpoint values, but it actually works for any continuous, differentiable function. Remember: MVT guarantees the instantaneous rate equals the average rate somewhere in the interval.
  19. Let qq be continuous on [−2,2][-2,2] and differentiable on (−2,2)(-2,2) with q(−2)=−6q(-2)=-6 and q(2)=2q(2)=2. Does MVT guarantee cc where q′(c)=2q'(c)=2?
    Answer: Yes, because qq is continuous on [−2,2][-2,2], differentiable on (−2,2)(-2,2), and q(2)−q(−2)2−(−2)=2−(−6)4=2\frac{q(2)-q(-2)}{2-(-2)}=\frac{2-(-6)}{4}=2. The function q meets both MVT criteria: continuous on [-2,2] and differentiable on (-2,2). The average rate of change is (q(2)-q(-2))/(2-(-2)) = (2-(-6))/4 = 8/4 = 2, which matches our target derivative value. Therefore, MVT guarantees at least one c in (-2,2) where q'(c) = 2. The fact that q starts negative and ends positive doesn't restrict the derivative to being negative; the function must increase overall with an average rate of 2. A common error is confusing the Intermediate Value Theorem (about function values) with MVT (about derivatives). For MVT problems, always calculate the average rate of change and compare it to the desired derivative value.
  20. Suppose pp is continuous on [−3,1][-3,1], differentiable on (−3,1)(-3,1), and p(−3)=4p(-3)=4, p(1)=0p(1)=0. Does MVT guarantee a cc?
    Answer: Yes, because pp is continuous on [−3,1][-3,1] and differentiable on (−3,1)(-3,1), so some cc satisfies p′(c)=−1p'(c)=-1. Function p satisfies both MVT requirements: continuous on [-3,1] and differentiable on (-3,1). The average rate of change is (0-4)/(1-(-3)) = -4/4 = -1. MVT guarantees there exists c in (-3,1) where p'(c) = -1. Choice C correctly states the conditions and conclusion. A common confusion is focusing on the function crossing the x-axis (p(1) = 0), which relates to IVT for finding roots, not MVT for finding derivative values. Strategy: Don't be distracted by special function values; MVT only depends on endpoint values and the two key conditions.

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